Complete formalization of the paper

For arbitrary universe levels $u,v$, the proposition consisting of the following thirteen assertions holds; the universally quantified groups and actions in distinct assertions are chosen independently, and the two existential examples are part of the same conjunction. (1) For every profinite group $G$ in universe $u$ and closed subgroups $N,H,K\le G$, assume that $N$ is normal in $G$ and pronilpotent, that either $G$ is prosupersolvable or $G/N$ is pronilpotent, and that every $g\in G$ can be expressed both as $g=nh$ with $n\in N,h\in H$ and as $g=n'k$ with $n'\in N,k\in K$. Then there exists $g\in G$ with $gHg^{-1}=K$ if and only if, for every natural prime $p$, there exist a Sylow pro-$p$ subgroup $P$ of $H$, a Sylow pro-$p$ subgroup $Q$ of $K$, and an element $g_p\in G$ with $g_pPg_p^{-1}=Q$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. Saying that a topological group $R$ is pronilpotent means that $R/U$ is nilpotent for every open normal subgroup $U$ of $R$, with the subgroup and quotient topologies understood. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. The product decompositions need not be unique, and $N\cap H$ and $N\cap K$ need not be trivial. The quantifier over primes includes primes absent from finite quotients of the subgroups; trivial groups and trivial Sylow subgroups are allowed, and the choices of $P,Q,g_p$ may depend on $p$. (2) For every profinite group $J$ in universe $u$, every finite nilpotent group $N$ in universe $v$ with the discrete topology, and every jointly continuous action of $J$ on $N$ by group automorphisms, assume that either $N\rtimes J$ is prosupersolvable or $J$ is pronilpotent. Here the semidirect product has multiplication $(n,j)(n',j')=(n(j\cdot n'),jj')$ and the product topology. Let $D$ be the set of natural primes $p$ for which $p$ divides the number of elements of $J/U$ for some open normal subgroup $U$ of $J$, and, for each $p\in D$, choose a subgroup $P_p\le J$ that is Sylow pro-$p$ in $J$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le J$ means a subgroup $P\le A$ that is closed in $J$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $J$ contained in $A$ with this quotient property. For $A\le J$, write $H^1(A,N)$ for the set of continuous maps $f:A\to N$ satisfying $f(xy)=f(x)(x\cdot f(y))$, modulo the equivalence relation $f\sim g$ if there is one $n\in N$ such that $g(x)=n^{-1}f(x)(x\cdot n)$ for every $x\in A$; its distinguished element is the class of the constant map $1$. Let $I_J(A,N)$ be the subset of $H^1(A,N)$ consisting of classes with a representative $f$ such that, for every $j\in J$, there exists $n_j\in N$ satisfying $j\cdot f(j^{-1}xj)=n_j^{-1}f(x)(x\cdot n_j)$ for every $x\in A$ for which $j^{-1}xj\in A$. This condition is imposed only on that intersection; it does not require $j$ to normalize $A$. The distinguished element of $I_J(A,N)$ is again the class of the constant map $1$. Then the simultaneous restriction map $H^1(J,N)\to\prod_{p\in D}I_J(P_p,N)$, sending $[f]$ to $([f|_{P_p}])_{p\in D}$, is both injective and surjective, and it sends the class of the constant map $1$ to the family of such classes. The target is the full product of these subsets, with no further compatibility condition between different primes. Saying that a topological group $R$ is pronilpotent means that $R/U$ is nilpotent for every open normal subgroup $U$ of $R$, with the subgroup and quotient topologies understood. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. Trivial $J$ or $N$ are allowed. If $D$ is empty, the product consists of the unique empty family, so the bijectivity assertion says that $H^1(J,N)$ has one element. (3) For every profinite group $G$ in universe $u$ and closed subgroups $N,J,H\le G$, assume that $N$ is normal in $G$, that every element of $G$ has a unique expression $nj$ with $n\in N$ and $j\in J$, that $N$ is pronilpotent, and that either $G$ is prosupersolvable or $J$ is pronilpotent. Assume also that $N\cap H$, regarded as a subgroup of $N$, is normal in $N$, and that for every natural prime $p$ there exist a Sylow pro-$p$ subgroup $P_p$ of $J$ and an element $g_p\in G$ with $g_pP_pg_p^{-1}\le H$. Then there exists one $g\in G$ such that $gJg^{-1}\le H$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. Saying that a topological group $R$ is pronilpotent means that $R/U$ is nilpotent for every open normal subgroup $U$ of $R$, with the subgroup and quotient topologies understood. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. The normality assumption on $N\cap H$ is in $N$, not a requirement of normality in $G$. The local witnesses can vary with $p$, whereas the conclusion concerns one conjugate of the whole subgroup $J$. All primes are quantified, including those with trivial Sylow subgroups; trivial groups and trivial factors in the unique product decomposition are permitted. (4) For every profinite group $G$ in universe $u$, every nonempty set $\Omega$ in universe $v$ with a $G$-action, and closed subgroups $N,J\le G$, assume that $N$ is normal in $G$, every element of $G$ has a unique expression $nj$ with $n\in N,j\in J$, $N$ is pronilpotent, and either $G$ is prosupersolvable or $J$ is pronilpotent. Assume that the action is transitive, meaning that for any $x,y\in\Omega$ some $g\in G$ satisfies $g\cdot x=y$; that the stabilizer $G_x=\{g\in G:g\cdot x=x\}$ is closed in $G$ for every $x\in\Omega$; and that there exists at least one $x\in\Omega$ for which $N\cap G_x$ is normal as a subgroup of $N$. Finally, assume that for every natural prime $p$ there exist a Sylow pro-$p$ subgroup $P_p$ of $J$ and a point $x_p\in\Omega$ fixed by every element of $P_p$. Then there exists a point $x\in\Omega$ fixed by every element of $J$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. Saying that a topological group $R$ is pronilpotent means that $R/U$ is nilpotent for every open normal subgroup $U$ of $R$, with the subgroup and quotient topologies understood. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. The points $x_p$ and the subgroups $P_p$ may depend on $p$, and they need not be related to the point witnessing the intersection-normality hypothesis. No topology on $\Omega$ is specified; the topological action hypothesis here is the closedness of every stabilizer. Singleton $\Omega$ and trivial groups are allowed, but empty $\Omega$ is excluded. All natural primes are included, even when their Sylow subgroups are trivial. (5) For every profinite group $J$ in universe $u$ and discrete topological group $N$ in universe $v$ equipped with a jointly continuous action of $J$ by group automorphisms, suppose every finite subset of $N$ generates a finite subgroup. Let $p,p_0$ be distinct natural primes, suppose that for every $n\in N$ there exists $k\in\mathbb N$ with $n^{p^k}=1$, and let $J_0,Q\le J$ be closed subgroups with $J_0$ normal in $J$. Assume that some $q_0\in Q$ has its integer powers dense in $Q$, and that every quotient $Q/U$ by an open normal subgroup $U$ of $Q$ has every element killed by some power of $p_0$. Let $f:J_0\to N$ be a continuous map satisfying $f(xy)=f(x)(x\cdot f(y))$ for all $x,y\in J_0$. Suppose that for every $q\in Q$ there exists $n_q\in N$ such that $q\cdot f(q^{-1}xq)=n_q^{-1}f(x)(x\cdot n_q)$ for every $x\in J_0$ with $q^{-1}xq\in J_0$. Then there exists a continuous map $g:J_0\to N$ satisfying $g(xy)=g(x)(x\cdot g(y))$ and all three following conditions: there is one $n\in N$ with $g(x)=n^{-1}f(x)(x\cdot n)$ for every $x\in J_0$; for every $q\in Q$ and $x\in J_0$ with $q^{-1}xq\in J_0$, one has $q\cdot g(q^{-1}xq)=g(x)$; and $g(q)=1$ for every $q\in Q\cap J_0$. Normality of $J_0$ ensures that the conjugate-membership condition in these identities always holds for $x\in J_0$. The finite-generation hypothesis permits $N$ to be infinite and includes the empty generating set; no uniform bound on the exponents $k$ is required. Trivial $N$, $J_0$, or $Q$ are permitted, a dense cyclic subgroup may be trivial, and the assumptions that $p,p_0$ are prime exclude $0$ and $1$. (6) For every profinite group $J$ in universe $u$ and discrete topological group $N$ in universe $v$ equipped with a jointly continuous action of $J$ by group automorphisms, suppose every finite subset of $N$ generates a finite subgroup. Let $p,p_0$ be distinct natural primes. Assume that every quotient $J/U$ by an open normal subgroup $U$ of $J$ is solvable and that every $n\in N$ satisfies $n^{p^k}=1$ for some $k\in\mathbb N$. Let $J_0$ be a closed normal subgroup of $J$ whose index is exactly $p_0$. For $A\le J$, write $H^1(A,N)$ for the set of continuous maps $f:A\to N$ satisfying $f(xy)=f(x)(x\cdot f(y))$, modulo the equivalence relation $f\sim g$ if there is one $n\in N$ such that $g(x)=n^{-1}f(x)(x\cdot n)$ for every $x\in A$; its distinguished element is the class of the constant map $1$. Let $I_J(A,N)$ be the subset of $H^1(A,N)$ consisting of classes with a representative $f$ such that, for every $j\in J$, there exists $n_j\in N$ satisfying $j\cdot f(j^{-1}xj)=n_j^{-1}f(x)(x\cdot n_j)$ for every $x\in A$ for which $j^{-1}xj\in A$. This condition is imposed only on that intersection; it does not require $j$ to normalize $A$. The distinguished element of $I_J(A,N)$ is again the class of the constant map $1$. Then restriction defines an injective and surjective map $H^1(J,N)\to I_J(J_0,N)$, $[f]\mapsto[f|_{J_0}]$, and sends the class of the constant map $1$ to that same distinguished class in the target. Since $J_0$ is normal, the conjugate-membership qualification in the definition of $I_J(J_0,N)$ holds for every $j\in J,x\in J_0$. The target consists of classes having the specified invariance property, not all classes on $J_0$. The group $N$ may be infinite or trivial, and the finite-generation hypothesis includes the empty finite set. The index is a natural-number cardinal, which is defined as $0$ at infinite index; its equality to the prime $p_0\ge2$ therefore requires finite index and excludes $J_0=J$. Both primes exclude $0$ and $1$. (7) For every profinite group $J$ in universe $u$, every finite group $N$ in universe $v$ with the discrete topology, and every jointly continuous action of $J$ on $N$ by group automorphisms, let $p$ be a natural prime and assume that every $n\in N$ satisfies $n^{p^k}=1$ for some $k\in\mathbb N$. Suppose that $N\rtimes J$, with multiplication $(n,j)(n',j')=(n(j\cdot n'),jj')$ and the product topology, is prosupersolvable. Let $Q\le J$ be closed, and assume that for every open normal subgroup $U$ of $J$, writing $\overline Q$ for the image of $Q$ under $J\to J/U$, every prime dividing $|\overline Q|$ is at most $p$, and every prime dividing the index $[J/U:\overline Q]$ is greater than $p$. For $A\le J$, write $H^1(A,N)$ for the set of continuous maps $f:A\to N$ satisfying $f(xy)=f(x)(x\cdot f(y))$, modulo the equivalence relation $f\sim g$ if there is one $n\in N$ such that $g(x)=n^{-1}f(x)(x\cdot n)$ for every $x\in A$; its distinguished element is the class of the constant map $1$. Then the restriction map $H^1(J,N)\to H^1(Q,N)$, $[f]\mapsto[f|_Q]$, is both injective and surjective, and sends the class of the constant map $1$ to that same distinguished class. Its target is the whole cohomology set on $Q$. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. The quotient groups $J/U$, their subgroup images, and their indices are finite here, so the natural-number cardinal and index conventions do not replace an infinite value by $0$. A subgroup image of order $1$ or index $1$ imposes no prime-divisor condition on that number. Trivial $J$ or $N$ are permitted; $p=0$ and $p=1$ are excluded by primality. (8) For every profinite group $G$ in universe $u$ and closed subgroups $N,H,K\le G$, assume that $N$ is a finite normal subgroup of $G$ and pronilpotent, and that either $G$ is prosupersolvable or $G/N$ is pronilpotent. Assume that every element of $G$ has a unique expression $nh$ with $n\in N,h\in H$ and also a unique expression $n'k$ with $n'\in N,k\in K$. Then there exists $g\in G$ with $gHg^{-1}=K$ if and only if, for every natural prime $p$, there exist a Sylow pro-$p$ subgroup $P$ of $H$, a Sylow pro-$p$ subgroup $Q$ of $K$, and $g_p\in G$ with $g_pPg_p^{-1}=Q$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. Saying that a topological group $R$ is pronilpotent means that $R/U$ is nilpotent for every open normal subgroup $U$ of $R$, with the subgroup and quotient topologies understood. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. The two unique-product assumptions include $N\cap H=N\cap K=\{1\}$. The groups $G,H,K$ need not be finite. Trivial $N$ and trivial other groups are permitted, all primes are included even when the relevant Sylow subgroups are trivial, and the local choices may vary with the prime. (9) For every profinite group $G$ in universe $u$ and closed subgroups $N,H,K\le G$, assume that $N$ is a finite normal subgroup of $G$ and pronilpotent, that either $G$ is prosupersolvable or $G/N$ is pronilpotent, and that every $g\in G$ can be expressed as $nh$ with $n\in N,h\in H$ and also as $n'k$ with $n'\in N,k\in K$. Then there exists $g\in G$ with $gHg^{-1}=K$ if and only if, for every natural prime $p$, there exist a Sylow pro-$p$ subgroup $P$ of $H$, a Sylow pro-$p$ subgroup $Q$ of $K$, and $g_p\in G$ with $g_pPg_p^{-1}=Q$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. Saying that a topological group $R$ is pronilpotent means that $R/U$ is nilpotent for every open normal subgroup $U$ of $R$, with the subgroup and quotient topologies understood. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Saying that $R$ is prosupersolvable means that every quotient $R/U$ by an open normal subgroup satisfies this series condition. The product decompositions are not required to be unique, and the intersections with $N$ are not required to be trivial. The groups $G,H,K$ need not be finite. Trivial groups are allowed, and every prime is quantified, including primes whose Sylow subgroups are trivial; the local choices can depend on $p$. (10) For every profinite group $G$ in universe $u$ and closed subgroups $N,H,K\le G$, suppose that $N$ is normal in $G$ and its multiplication is commutative, and that every element of $G$ can be expressed both as $nh$ with $n\in N,h\in H$ and as $n'k$ with $n'\in N,k\in K$. Suppose also that for every natural prime $p$ there exist a Sylow pro-$p$ subgroup $P_p$ of $K$ and an element $g_p\in G$ with $g_pP_pg_p^{-1}\le H$. Then there exists a single $g\in G$ with $gKg^{-1}\le H$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. The product expressions need not be unique, and neither intersection with $N$ is required to be trivial. The conclusion is containment rather than equality. Neither the local subgroup nor its conjugating element must be independent of $p$. All natural primes are included; trivial groups, trivial $N$, and trivial Sylow subgroups are permitted, and none of these groups is required to be finite. (11) For every profinite group $G$ in universe $u$, every nonempty set $\Omega$ in universe $v$ with a $G$-action, and closed subgroups $N,H\le G$, suppose that $N$ is normal in $G$ and has commutative multiplication, and that every element of $G$ can be expressed as $nh$ with $n\in N,h\in H$. Suppose that the action is transitive, meaning that for every $x,y\in\Omega$ some $g\in G$ satisfies $g\cdot x=y$, and that each stabilizer $G_x=\{g\in G:g\cdot x=x\}$ is closed in $G$. If for every natural prime $p$ there exist a Sylow pro-$p$ subgroup $P_p$ of $H$ and a point $x_p\in\Omega$ fixed by all elements of $P_p$, then there exists a point $x\in\Omega$ fixed by every element of $H$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. The points and subgroups in the hypothesis may vary with $p$. No topology on $\Omega$ is specified, and there is no separately assumed continuity of its action beyond the closed-stabilizer condition. Empty $\Omega$ is excluded, but singleton $\Omega$, trivial groups, and trivial Sylow subgroups are allowed. All primes are quantified, no finiteness of $\Omega$ or $G$ is required, and the expression $nh$ need not be unique. (12) There exists a group homomorphism $a:S_3\to\operatorname{Aut}(Q_8)$, where $S_3$ is the permutation group of the three-element set $\{0,1,2\}$ and $Q_8$ is the quaternion group of order $8$, with the following simultaneous properties. The semidirect product $E=Q_8\rtimes_a S_3$, whose multiplication is $(n,s)(n',s')=(n\,a(s)(n'),ss')$, is isomorphic as an abstract group to the group of invertible $2\times2$ matrices over $\mathbb Z/3\mathbb Z$. Let $H^1_b(A,Q_8)$, for an action $b:A\to\operatorname{Aut}(Q_8)$, denote the quotient of all maps $f:A\to Q_8$ satisfying $f(xy)=f(x)b(x)(f(y))$ by the equivalence relation $f\sim g$ when there exists one $n\in Q_8$ with $g(x)=n^{-1}f(x)b(x)(n)$ for every $x\in A$; no continuity is imposed in this definition. Then $H^1_a(S_3,Q_8)$ has exactly two elements, while for every natural prime $p$ and every Sylow $p$-subgroup $P$ of $S_3$, any two elements of $H^1_{a|_P}(P,Q_8)$ are equal. In addition, writing $B=\{(n,1):n\in Q_8\}$ and $C=\{(1,s):s\in S_3\}$, there exists a subgroup $J'\le E$ such that every element of $E$ has a unique expression $bj'$ with $b\in B,j'\in J'$, for every natural prime $p$ there exist Sylow $p$-subgroups $P_p\le C$, $Q_p\le J'$ and $e_p\in E$ with $e_pP_pe_p^{-1}=Q_p$, and there is no $e\in E$ with $eCe^{-1}=J'$. Here a Sylow $p$-subgroup is a subgroup maximal among those whose every element is killed by some power of $p$. The quantifiers include primes other than $2$ and $3$, whose Sylow subgroups in $S_3$ are trivial. Every cohomology set just defined contains the class of the constant map $1$, so the assertion that any two restricted classes are equal is not an assertion about an empty set. The cardinality-two assertion concerns a finite set and therefore does not use the convention assigning natural-number cardinal $0$ to an infinite set. The isomorphism with the matrix group is asserted to exist, and no particular choice of it or of the action $a$ is specified. (13) There exist a profinite group $G$ whose underlying type lies in universe $0$ and subgroups $N,J,H\le G$ with all the following properties. The group $G$ is finite and, as an abstract group, is isomorphic to $(C_3^{\{0,1,2\}})\rtimes S_3$, where $C_3$ is the additive group of $\mathbb Z/3\mathbb Z$ written multiplicatively, the function group has pointwise multiplication, $S_3$ permutes $\{0,1,2\}$, and the action on functions is $(s\cdot b)(i)=b(s^{-1}(i))$; the semidirect multiplication is $(b,s)(b',s')=(b(s\cdot b'),ss')$. The subgroup $N$ is isomorphic to the group of triples $(a,b,c)\in(\mathbb Z/3\mathbb Z)^3$ with multiplication $(a,b,c)(a',b',c')=(a+a',b+b',c+c'+ab')$, identity $(0,0,0)$, and inverse $(-a,-b,-c+ab)$. The subgroup $J$ is isomorphic to the additive group of $\mathbb Z/6\mathbb Z$ written multiplicatively, and $H$ is isomorphic to $C_3\times S_3$. Their cardinalities are exactly $|G|=162$, $|N|=27$, $|J|=6$, and $|H|=18$. The subgroup $N$ is normal in $G$, every element of $G$ has a unique expression $nj$ with $n\in N,j\in J$, both $N$ and $J$ are nilpotent, and $G$ satisfies the following series condition. For a group $R$, the series condition used here means that there exist $m\in\mathbb N$ and a nondecreasing sequence $(S_i)_{i\in\mathbb N}$ of normal subgroups of $R$ such that $S_0=\{1\}$, $S_m=R$, and, for each $i<m$, some $r_i\in R$ satisfies $S_{i+1}=\langle S_i,r_i\rangle$. Repeated terms are allowed, and $m=0$ is allowed precisely when $R$ is trivial. Each of $N,J,H$ is closed in $G$. For every natural prime $p$ there exist a Sylow pro-$p$ subgroup $P_p$ of $J$ and an element $g_p\in G$ with $g_pP_pg_p^{-1}\le H$, yet there is no $g\in G$ with $gJg^{-1}\le H$, and $N\cap H$ is not normal as a subgroup of $N$. A Sylow pro-$p$ subgroup $P$ of a subgroup $A\le G$ means a subgroup $P\le A$ that is closed in $G$, for which every quotient $P/U$ by an open normal subgroup of $P$ has the property that every element is killed by some power $p^k$ with $k\in\mathbb N$, and that is maximal under inclusion among the closed subgroups of $G$ contained in $A$ with this quotient property. The local witnesses may depend on the prime, and the prime quantifier includes primes other than $2$ and $3$, with trivial Sylow subgroups of $J$. All isomorphisms asserted are group isomorphisms, with no specified compatibility among them. The explicit finite cardinalities exclude trivial groups in this existential assertion, and none of these cardinalities uses the infinite-cardinality value $0$.